Let f(x) = x^2 + 17x + 72. What are the zeros of the function? Enter your answers in the boxes
The Correct Answer and Explanation is:
First, let’s find the zeros of the quadratic function
f(x)=x2+17x+72.f(x) = x^2 + 17x + 72.f(x)=x2+17x+72.
The zeros occur where f(x)=0f(x) = 0f(x)=0, so we solve the equation
x2+17x+72=0.x^2 + 17x + 72 = 0.x2+17x+72=0.
We can use the quadratic formula x=−b±b2−4ac2a,x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a},x=2a−b±b2−4ac,
where a=1a = 1a=1, b=17b = 17b=17, and c=72c = 72c=72.
First, compute the discriminant: b2−4ac=172−4⋅1⋅72=289−288=1.b^2 – 4ac = 17^2 – 4 \cdot 1 \cdot 72 = 289 – 288 = 1.b2−4ac=172−4⋅1⋅72=289−288=1.
Since the discriminant is positive, we have two real solutions: x=−17+12=−162=−8,x = \frac{-17 + 1}{2} = \frac{-16}{2} = -8,x=2−17+1=2−16=−8, x=−17−12=−182=−9.x = \frac{-17 – 1}{2} = \frac{-18}{2} = -9.x=2−17−1=2−18=−9.
So the zeros of the function are −8and−9.\boxed{-8} \quad \text{and} \quad \boxed{-9}.−8and−9.
Here is a detailed explanation:
A quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 has solutions given by the quadratic formula. The discriminant b2−4acb^2 – 4acb2−4ac tells us about the nature of the roots. If it is positive, there are two distinct real solutions; if it is zero, there is one real repeated solution; and if it is negative, there are two complex solutions.
For this function, the discriminant was positive, so we expected two real and distinct zeros. Calculating it carefully gave us a discriminant of 1, a perfect square, which means our solutions are rational numbers.
We could also have factored directly because we are looking for two numbers that multiply to 727272 (the constant term) and add up to 171717 (the coefficient of xxx). These two numbers are 888 and 999, and because both signs are positive in the quadratic but the constant term is positive, both factors in the equation would be negative when written as factors: (x+8)(x+9)=0,(x + 8)(x + 9) = 0,(x+8)(x+9)=0,
leading to solutions x=−8x = -8x=−8 and x=−9x = -9x=−9. This matches exactly with what we obtained using the quadratic formula. This shows that factoring and the quadratic formula are consistent methods for solving quadratic equations
